PUC Science इयत्ता ११ - Karnataka Board PUC Question Bank Solutions for Mathematics

\[\lim_{x \to 3} \left( \frac{1}{x - 3} - \frac{3}{x^2 - 3x} \right)\] 

\[\lim_{x \to 1} \left( \frac{1}{x - 1} - \frac{2}{x^2 - 1} \right)\]

\[\lim_{x \to 3} \left( x^2 - 9 \right) \left[ \frac{1}{x + 3} + \frac{1}{x - 3} \right]\] 

\[\lim_{x \to 1} \frac{x^4 - 3 x^3 + 2}{x^3 - 5 x^2 + 3x + 1}\] 

\[\lim_{x \to 2} \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6}\] 

\[\lim_{x \to 1} \frac{1 - x^{- 1/3}}{1 - x^{- 2/3}}\] 

\[\lim_{x \to 3} \frac{x^2 - x - 6}{x^3 - 3 x^2 + x - 3}\]

\[\lim_{x \to - 2} \frac{x^3 + x^2 + 4x + 12}{x^3 - 3x + 2}\]

\[\lim_{x \to 1} \frac{x^3 + 3 x^2 - 6x + 2}{x^3 + 3 x^2 - 3x - 1}\]

\[\lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x^3 - 3 x^2 + 2x} \right]\] 

\[\lim_{x \to 1} \frac{\sqrt{x^2 - 1} + \sqrt{x - 1}}{\sqrt{x^2 - 1}}, x > 1\] 

\[\lim_{x \to 1} \left\{ \frac{x - 2}{x^2 - x} - \frac{1}{x^3 - 3 x^2 + 2x} \right\}\] 

Evaluate the following limit:

\[\lim_{x \to 1} \frac{x^7 - 2 x^5 + 1}{x^3 - 3 x^2 + 2}\] 

\[\lim_{x \to a} \frac{\left( x + 2 \right)^{5/2} - \left( a + 2 \right)^{5/2}}{x - a}\] 

\[\lim_{x \to a} \frac{\left( x + 2 \right)^{3/2} - \left( a + 2 \right)^{3/2}}{x -  a}\]

\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\] 

\[\lim_{x \to a} \frac{x^{2/7} - a^{2/7}}{x - a}\] 

\[\lim_{x \to a} \frac{x^{5/7} - a^{5/7}}{x^{2/7} - a^{2/7}}\] 

\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]

\[\lim_{x \to 27} \frac{\left( x^{1/3} + 3 \right) \left( x^{1/3} - 3 \right)}{x - 27}\]